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Solids of Revolution: The Washer Method

 

Introduction

Imagine for a moment that you work for a chocolate factory and help to create new treats. You’re very good at your job and have developed a design for a new candy consisting of a chocolate shell with a caramel-filled cavity. You want the outside of the shell to be 1.5 inches tall and have the same curve as the equation \( f(x)=- \frac{1}{9}\ x^2+ \frac{3}{4}\ \). You also need there to be an opening in the center for the filling, which you want to have the same curve as \( g(x)= \frac{1}{2}\ - \frac{1}{2}\ e^{-x} \). There’s only one problem: before you can begin manufacturing your masterpiece, your boss wants to know what the volume of this candy is so he can order the correct amount of chocolate. Is there a way to calculate the volume of such a complicated shape? Fortunately, there is, with a little help from calculus.

What is the Washer Method?

As discussed above, we know the wall of our solid will consist of the space between the following two curves. Since it must be 1.5 inches tall, we will only look at the interval \( 0 \leq x \leq 1.5 \):

Imagine rotating the area between these curves around the x-axis – that would give us the solid we are trying to create! The washer method is a way to calculate the volume of such a solid by dividing it into disks (or washers in this case, since there is a hole in the middle) and adding up the volumes of those disks. First we'll use the washer method to get an estimate of our shape’s volume by dividing our solid into 10 washers. Since our interval is 1.5 inches long, this means that each washer will have the following height: \[ h=1.5 \div 10=0.15 \ \textrm{inches} \]

For our first washer, we start at \( x=0 \). At this point, the distance from the center to the outer wall (in other words, the outer radius) is: \( f(x)=- \frac{1}{9}\ (0)^2+ \frac{3}{4}\ = \frac{3}{4}\ \) inches wide. Similarly, the hole at the center has a radius of \( g(x)= \frac{1}{2}\ - \frac{1}{2}\ e^0=0 \) inches. This means our first washer is a disk of radius 0.75 inches with a hole of radius 0 inches. This is a cylinder we can easily compute the volume of: \[ V_1=( \pi (0.75)^2 \cdot 0.15)-( \pi (0)^2 \cdot 0.15)=0.265 \ \textrm{in}^2 \]

That’s one washer done. By following the same steps, we can compute the volume for all 10 washers along our interval:

By creating these washers, we’ve developed an approximation of our solid that looks like the model below. The model is 1.5 inches tall at its tallest point and 1.5 inches wide at its widest point:

Now we can simply add these up to get the total volume! Doing this, we get \( 1.898 \ \textrm{inches}^2 \). Remember, however, that this is not the exact volume because the edges of the solid above are not smooth exacltly like our equations- it is an approximation.

Applying the Washer Method

So far, we have an estimate of our solid’s volume. We know that the more washers we use to calculate the volume, the more accurate our calculations will be since they will follow the curve of our equations more closely. As you know, integrating a function over an interval is like adding an infinite number of rectangles to calculate the area beneath the function. Thus, we can do the same thing for our solid by using integration in our equation for the volume of a washer.

Therefore, the equation we used earlier to calculate the volume of one washer \[ V= \pi R^2h- \pi r^2h= \pi h(R^2-r^2) \] becomes \[ V= \pi \int_{a}^{b} (R^2-r^2) \,dx \] where \( R \) represents the radius for the outer edge and \( r \) represents the radius for the inner cavity on the interval \( a \leq x \leq b \) .

Using this equation on our example from before, we get the following integral: \[ \pi \int_{0}^{1.5} (f(x)^2-g(x)^2) \,dx = \pi \int_{0}^{1.5} \Big( (- \frac{1}{9}\ x^2+ \frac{3}{4}\ )^2-( \frac{1}{2}\ - \frac{1}{2}\ e^{-x})^2 \Big) \,dx \] All that’s left to do is evaluate the integral. \[ \pi \int_{0}^{1.5} \Big( \frac{1}{81}\ x^3- \frac{1}{6}\ x^2+ \frac{9}{16}\ -( \frac{1}{4}\ - \frac{1}{2}\ e^{-x}+ \frac{1}{4}\ e^{-2x}) \Big) \,dx \] \[ = \pi \int_{0}^{1.5} \Big( \frac{1}{81}\ x^3- \frac{1}{6}\ x^2+ \frac{5}{16}\ + \frac{1}{2}\ e^{-x}- \frac{1}{4}\ e^{-2x} \Big) \,dx \] \[ \pi \Big[ \frac{1}{405}\ x^4- \frac{1}{18}\ x^3+ \frac{5}{16}\ x- \frac{1}{2}\ e^{-x}+ \frac{1}{8}\ e^{-2x} \Big] \] \[ \pi \bigg[ \frac{(1.5)^4}{405}\ - \frac{(1.5)^3}{18}\ + \frac{5(1.5)}{16}\ - \frac{1}{2}\ e^{-1.5}+ \frac{1}{8}\ e^{-3}- \Big( \frac{0^4}{405}\ - \frac{0^3}{18}\ + \frac{5(0)}{16}\ - \frac{1}{2}\ e^0+ \frac{1}{8}\ e^0 \Big) \bigg] \] \[ =1.7896 \]

Therefore, we have determined that the volume of our solid is \( 1.7896 \ \textrm{in}^2 \) and can tell our boss exactly what volume of chocolate we need to make our candy.

Why These Functions?

The functions used in the preceding example were chosen for three primary reasons. The first and most salient of these reasons is the functions have interesting behavior that reinforces the learner’s understanding of the material. Compared to linear functions, the curving path of these functions allow the learner to observe the relationship between the functions and the changing widths of the washers on both the inside and outside of the model. Furthermore, both of the functions have steeper slopes at some point along the integral (for \( f(x) \) this increases with \( x \) and for \( g(x) \) this decreases with \( x \) ) which allows the learner to observe the relationship between the slope and the changes in the sizes of the washers. Secondly, these functions together produce a shape that students could reasonably imagine being a chocolate candy. This makes the material more approachable by providing a model of something that could (with a little imagination) serve a real-world purpose. Lastly, these functions result in an equation that is easy to integrate, making the example accessible to students who don’t remember (or struggled with) more advanced integration techniques. Thus, the example uses math that is not trivial while still keeping the focus on the new method.

Author: Sarah Bombrys

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